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Given that $x^4 + ax^3 + 3x^2 +bx +1$ is always greater than equal to $0$ for all $x$ belongs to $\mathbb R$, find $\max(a^2 + b^2)$.

What I did was to show that that above expression is equivalent to $$[x(x+a/2)]^2 + [(12-a^2)(x + 2b/(12-a^2))^2]/4 + (12-a^2 - b^2) /(12-a^2)\geq 0,$$ from this one case is that if $ 12-a^2\geq 0$, then the expression trivially is greater/equal zero, which given max of $a^2+b^2$ to be $12$, but what about if $12-a^2<=0$, then is max of $a^2+ b^2$ greater than 12 possible and what is the exact maximum ? Conclusion: there are two such $f(x)$ for which maxima is attained those are $f(x) = x^4 -2√5x^3 + 3x^2+2√5x +1$ and other as said by achille hui.

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    $\begingroup$ I will do that from next time $\endgroup$ Aug 29, 2021 at 10:33
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    $\begingroup$ The maximum is at least $40$ because $x^4 + 2\sqrt{5}x^3 + 3x^2 - 2\sqrt{5}x + 1 = (x^2 + \sqrt{5}x - 1)^2$ is non-negative for all $x$. $\endgroup$ Aug 29, 2021 at 10:44
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    $\begingroup$ How to prove it , as such answer given was 40 ,i want to know how one gets that maximum from that 12-a^2<=0 case in my expression i got , for 12-a^2>=0 i got max to be 12 , can u show for the <=0 case max to be 40? $\endgroup$ Aug 29, 2021 at 10:47
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    $\begingroup$ I didn't have a direct proof that $40$ is maximum. I get that configuration by computing the resultant of the polynomial and its derivative. That will give us the boundary ( in $(a,b)$ parameter space) where the condition "non-negative for all $x$" start to fail. $\endgroup$ Aug 29, 2021 at 10:52
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    $\begingroup$ I tried this: As the equation has no real root the roots will be pairwise complex conjugate of each other. Also, their magnitudes will be pairwise multiplicative inverses of each other. Then using Vieta's theorem, we have to find maximum value. But I couldn't manage to get the max value. $\endgroup$
    – Ilovemath
    Aug 29, 2021 at 11:11

2 Answers 2

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This image contains the solution to your question I have started using MathJax but am not fluent in using it so posting the solution like this.

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  • $\begingroup$ I am learning to use MathJax but typing this solution is very hectic as I am not fluent at it. Sorry for that. $\endgroup$
    – Ilovemath
    Aug 29, 2021 at 12:07
  • $\begingroup$ Really gud way to solve it , i dont mind non latex sol , but do u think in my way , for 12-a^2<= 0 case , what could be the approach after that modified expression i got ? Among the three terms in my expression first and third will be positive and the middle one would be negative , so what we need to do to ensure first + third sum always greater than or equal to middle term? $\endgroup$ Aug 29, 2021 at 12:58
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    $\begingroup$ If you take cosθ=1 and cosϕ=−1 , you get a2=20 meaning that the maximum would be achieved when 12−a2<0 but the sum of 1st and 3rd quantities as well as the 2nd quantity are functions of x.So when you try to minimise the sum of 1st and 3rd, you don't know whether at that value the 2nd quantity is minimum so that you can make their sum positive to obtain a boundary condition in a and b. That is keeping us away from the correct answer. I think. – $\endgroup$
    – Ilovemath
    Aug 29, 2021 at 13:40
  • $\begingroup$ I see so this is the best way to solve the problem ( ur method )? My method will need much much work later on for simplification u mean for boundary condition? $\endgroup$ Aug 29, 2021 at 13:45
  • $\begingroup$ One doubt regarding your solution also @ilove math that initial expression can be = 0 too for some x , that means there can exists real roots too isnt ? So why take complex roots from start ? $\endgroup$ Aug 29, 2021 at 13:58
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Remark: According to @achille hui's comment $x^4 + 2\sqrt{5}x^3 + 3x^2 - 2\sqrt{5}x + 1 = (x^2 + \sqrt{5}\, x - 1)^2$, we give the following solution:


Let $f(x) = x^4 + ax^3 + 3x^2 + bx + 1$.

Fact 1: If $f(x)\ge 0$ for all $x\in \mathbb{R}$, then $a^2 + b^2 \le 40$.
(The proof is given at the end.)

By Fact 1, we have $a^2 + b^2 \le 40$. On the other hand, if $a = 2\sqrt{5}, ~ b = -2\sqrt5$ with $a^2 + b^2 = 40$, we have $x^4 + 2\sqrt{5}\, x^3 + 3x^2 - 2\sqrt{5}\, x + 1 = (x^2 + \sqrt{5}x - 1)^2 \ge 0$ for all $x\in \mathbb{R}$.

Thus, the maximum of $a^2 + b^2$ is $40$.

We are done.


Proof of Fact 1: We split into two cases:

Case 1: If $ab \ge 0$, since $f(1) = a + b + 5 \ge 0$ and $f(-1) = - (a + b) + 5 \ge 0$, we have $-5 \le a + b \le 5$ and thus $a^2 + b^2 \le (a + b)^2 \le 25 < 40$.

Case 2: If $ab < 0$, WLOG, assume that $a > 0, ~ b < 0$ (otherwise, $a \to -a, ~ b \to -b, ~ x \to - x$).

Let $x_1 = \frac{-\sqrt5 + 3}{2} > 0$ and $x_2 = \frac{-\sqrt5 - 3}{2} < 0$.
(Note: $x_1$ and $x_2$ are the roots of $x^2 + \sqrt{5}\, x - 1 = 0$. See @achille hui's comment.)

From $f(x_1)\ge 0$, we have $$b \ge -x_1^3 - ax_1^2 - 3x_1 - \frac{1}{x_1}. \tag{1}$$

From $f(x_2)\ge 0$, we have $$b \le -x_2^3 - ax_2^2 - 3x_2 - \frac{1}{x_2}. \tag{2}$$

From (1) and (2), we have $$-x_1^3 - ax_1^2 - 3x_1 - \frac{1}{x_1} \le -x_2^3 - ax_2^2 - 3x_2 - \frac{1}{x_2}$$ which results in (note: $x_1 + x_2 = -\sqrt5, x_1x_2 = -1, x_1 - x_2 = 3$) $$a \le \frac{x_1^3 - x_2^3 + 3x_1 - 3x_2 + \frac{1}{x_1} - \frac{1}{x_2}}{x_2^2 - x_1^2} = \frac{x_1^2 + x_1x_2 + x_2^2 + 3 - \frac{1}{x_1x_2}}{-(x_1 + x_2)} = 2\sqrt{5}.$$

Then, using $-x_1^3 - ax_1^2 - 3x_1 - \frac{1}{x_1} \le b < 0$ and $0 < a \le 2\sqrt5$, we have \begin{align*} a^2 + b^2 &\le a^2 + \left(-x_1^3 - ax_1^2 - 3x_1 - \frac{1}{x_1}\right)^2\\ &\le (2\sqrt{5})^2 + \left(x_1^3 + 2\sqrt{5}\cdot x_1^2 + 3x_1 + \frac{1}{x_1}\right)^2\\ &= 40. \end{align*} Note: Here, we may use $x_1^2 + \sqrt{5}\, x_1 - 1 = 0$ to reduce the calculations. For example, $x_1^3 = -\sqrt{5}\, x_1^2 + x_1$.

We are done.

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  • $\begingroup$ One thing i didnt quite get how u thought of getting f(3-√5/2) AND f(-3-√5/2 ) for case 2?not any other real numbers? $\endgroup$ Aug 30, 2021 at 6:48
  • $\begingroup$ And achille hui comment uses derivative too for getting boundary conditions on a and b isnt ? In yours derivative wasnt used $\endgroup$ Aug 30, 2021 at 6:51
  • $\begingroup$ @Paracetamol $\frac{-\sqrt5 + 3}{2}$ and $\frac{-\sqrt5 - 3}{2}$ are the two roots of $x^2 + \sqrt{5}x - 1 = 0$. $\endgroup$
    – River Li
    Aug 30, 2021 at 8:00
  • $\begingroup$ @Paracetamol You can guess, at maximum of $a^2+b^2$, it should hold $x^4 + ax^3 + 3x^2 +bx +1 = (x^2 + px + q)^2$. If you guess that, you get $a = 2\sqrt{5}, b = -2\sqrt{5}$ etc. Although it is a guess, it is enough to motivate the solution. $\endgroup$
    – River Li
    Aug 30, 2021 at 8:05
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    $\begingroup$ @Paracetamol If we know achille hui's result $x^4 + 2\sqrt{5}x^3 + 3x^2 - 2\sqrt{5}x + 1 = (x^2 + \sqrt{5}x - 1)^2$, we just find the two roots of $x^2 + \sqrt{5}x - 1 = 0$, say $x_1, x_2$. We have $f(x_1) = f(x_2) = 0$ in his case $a = 2\sqrt5, b = 2\sqrt5$. For general $a, b$, we can go from $f(x_1) \ge 0$ and $f(x_2) \ge 0$ which gives conditions for $a, b$. If we don't know achille hui's result, I will guess, when $a^2 + b^2$ is maximized, it should admit $x^4 + ax^3 + 3x^2 +bx +1 = (x^2 + px + q)^2$ which actually gives $a = 2\sqrt5, b= -2\sqrt5$. $\endgroup$
    – River Li
    Aug 30, 2021 at 8:29

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